# Induction proof of the area of a square

English is not my first language, so I'm sorry if I'm not very clear. I can clarify any question you have. Also, I don't know how to use that math formatting so I apologize for it.

I was asked to come up with a formula from the following image:

And the "hint" was: $1 + 2\times2^2 + 3\times3^2 + 4\times4^2+ 5\times5^2= \cdots$

It seems like a way to calculate the area of a square:

$$k^2 = 1 + 2\times2^2 + \cdots + l \times l^2$$

Where $1 + 2 + \cdots + l = k$, and $k$ is the side of the square

I tested this with some numbers:

\begin{align} 10^2 &= 1 + 2 \times 2^2 + 3 \times 3^2 + 4 \times 4^2\\ 100 &= 1 + 8 + 27 + 64\\ 100 &= 100\\ \end{align}

I have to prove this using induction so I tried this:

$$(k+n)^2 = 1 + 2 \times 2^2 + \cdots + l \times l^2 + n \times n^2$$

If I understood induction correctly, I plug the first formula inside the second one:

$$(k+n)^2 = k^2 + n \times n^2$$

…and this is where I'm stuck. I can't figure out how to make these two equal, even though if I substitute them with numbers it appears the formula is correct.

• $$\sum_{k=1}^nk*k^2=\frac{n^2(n+1)^2}4$$ Sep 17 '14 at 19:05
• For some basic information about writing math at this site see e.g. here, here, here and here. Sep 17 '14 at 19:08
• @DietrichBurde What do you mean? k^2 = 1 + 2*2^2 + ... + l * l^2 This exact line appears in the right hand side doesn't it? Sep 17 '14 at 19:13
• @Darksonn Thank you very much, I'll try to familiarize myself with it Sep 17 '14 at 19:15
• @DietrichBurde But can't I plug from the first equation into the second one? If k^2 = 1 + 2*2^2 + ... + l * l^2, and 1 + 2*2^2 + ... + l * l^2 appears in the second last equation, can't I substitute the whole thing for k^2? That was my understanding on how induction worked... Sep 17 '14 at 19:20

You want to show that $$k(n)^2=1\cdot1^2+2\cdot2^2+3\cdot3^2+\ldots+n\cdot n^2,$$ where $k(n)=1+2+\cdots+n$, for any $n$. For $n=1$, you have $k(1)^2=1^2=1$ on the left and $1\cdot 1^2=1$ on the right. Now assuming the equation holds for $n$, you want to show that it also holds for $n+1$. Note that $k(n+1)=k(n)+n+1$. Now $$k(n+1)^2=\left(k(n)+n+1\right)^2=k(n)^2+(n+1)\left(2 k(n)+n+1\right)\\=1\cdot 1^2 + 2\cdot 2^2 + \ldots + n\cdot n^2+(n+1)\left(2 k(n)+n+1\right).$$ For this to help you, you need to also show that $2k(n)+n+1=(n+1)^2$, or that $k(n)=\frac{1}{2}(n+1)^2-\frac{1}{2}(n+1)=\frac{1}{2}n(n+1).$ (You can show that by induction, too, if you like.) Once that step is done, you'll have shown that the equation holds for $n=1$, and that if it holds for some $n$ it also holds for $n+1$; and you can conclude by induction that it holds for all $n\ge 1$.
• Sorry, I don't understand this very well... Would you mind clarifying it a bit more? First thing: why do I want to show $$k(n)^2=1\cdot1^2+2\cdot2^2+3\cdot3^2+\ldots+n\cdot n^2?$$ Is what I'm trying to prove not $$k^2=1\cdot1^2+2\cdot2^2+3\cdot3^2+\ldots+n\cdot n^2$$ Why $k(n)^2$ instead of $k^2$? Sep 17 '14 at 19:33
• I wrote $k(n)$ because $k$ is a function of $n$, defined by $k(n)=1+2+\cdots+n$. Just writing "$k$" without showing the dependence on $n$, especially when more than one value of $n$ will be considered, can be confusing. Sep 18 '14 at 13:32